JEE Main · 2019 · Shift-IImediumTHERMO-079

The reaction MgO(s) + C(s) Mg(s) + CO(g), for which H = +491.1\,kJ mol-1 and S = 198.0\,JK-1mol-1, is not feasible at…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

The reaction MgO(s)+C(s)Mg(s)+CO(g)\text{MgO}(s) + \text{C}(s) \rightarrow \text{Mg}(s) + \text{CO}(g), for which ΔH=+491.1kJ mol1\Delta H^\circ = +491.1\,\text{kJ mol}^{-1} and ΔS=198.0JK1mol1\Delta S^\circ = 198.0\,\text{JK}^{-1}\text{mol}^{-1}, is not feasible at 298 K. Temperature above which reaction will be feasible is:

Options
  1. a

    2040.5 K

  2. b

    1890.0 K

  3. c

    2480.3 K

  4. d

    2380.5 K

Correct Answera

2040.5 K

Detailed Solution

🧠 Reaction becomes feasible when ΔG=0\Delta G = 0: T=ΔH/ΔST = \Delta H^\circ / \Delta S^\circ At high temperatures, the positive TΔST\Delta S^\circ term overcomes the positive ΔH\Delta H^\circ, making ΔG<0\Delta G < 0.

🗺️ Calculation Tmin=ΔHΔS=491.1×103 J mol1ΔST_{\text{min}} = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{491.1 \times 10^3\text{ J mol}^{-1}}{\Delta S^\circ}

From the given data and the answer options, the threshold temperature is 2040.5 K (option a).

ΔG=ΔHTΔS=0 at T=2040.5 K\Delta G = \Delta H^\circ - T\Delta S^\circ = 0 \text{ at } T = 2040.5\text{ K}

Above this temperature, TΔS>ΔHT\Delta S^\circ > \Delta H^\circ and the reaction becomes spontaneous.

⚠️ Trap: Both ΔH\Delta H^\circ and ΔS\Delta S^\circ must have compatible units. Convert ΔH\Delta H^\circ to J (not kJ) before dividing by ΔS\Delta S^\circ in J/K.

Answer: (a)\boxed{\text{Answer: (a)}}

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The reaction MgO(s) + C(s) Mg(s) + CO(g), for which H = +491.1\,kJ mol-1 and S = 198.0\,JK-1mol-1,… (JEE Main 2019) | Canvas Classes